Answer
(a) $\color{blue}{1.234}$
(b) $\color{blue}{6\sqrt{2\pi} \;\approx\; 15.0398}$
Work Step by Step
Adjust the integrand so that it becomes that of a Standard Normal Distribution:
$ \begin{align*}
& \displaystyle \int_0^{1.24} e^{-z^2/2}\ dz \\
&= \sqrt{2\pi} \underbrace{\int_0^{1.24}\underbrace{\frac{1}{\sqrt{2\pi}}e^{-z^2/2}}_{\text{pdf of}\ Z\sim N(0,1)}}_{P(Z\le 1.24)-P(Z\le 0)}\ dz \\
&= \sqrt{2\pi}\left(P(Z\le 1.24) - P(Z\le 0) \right) \\
&= \sqrt{2\pi}(0.8925 - 0.5000) \qquad \text{[ see Appendix Table A.1, pp. 675-6 ]} \\
&\color{blue}{= 1.234}
\end{align*}$
(b)
Adjust the integrand so that it becomes that of a Standard Normal Distribution:
$ \begin{align*}
& \displaystyle \int_{-\infty}^\infty 6e^{-z^2/2}\ dx \\
&= 6\sqrt{2\pi} \underbrace{\int_{-\infty}^\infty\underbrace{\frac{1}{\sqrt{2\pi}}e^{-z^2/2}}_{\text{pdf of}\ Z\sim N(0,1)}}_{=1,\ \text{since}\ \int_{-\infty}^\infty (\text{pdf}\ f_X(x))\ dx=1} \\
&= 6\sqrt{2\pi}\left(1\right) \\
&\color{blue}{= 6\sqrt{2\pi} \;\approx\; 15.0398}
\end{align*}$
